For $f(z)=z^m$,
$$
i\,df\wedge d\bar f=2m^2r^{2m-2}r\,dr\wedge d\theta.
$$
The pullback area is therefore
$$
\int_{\mathbb C}\frac{i\,df\wedge d\bar f}{(1+|f|^2)^2}
=4\pi m^2\int_0^\infty\frac{r^{2m-1}}{(1+r^{2m})^2}\,dr
=2\pi m.
$$
The same form integrates to $2\pi$ on the target sphere, so
$$
\boxed{\deg f=m}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 313 / 3 / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-313/3/ii/solution